When ammonia is mixed with hydrogen chloride (HCl), the white solid ammonium chloride (NH4Cl) is produced. Suppose 10.0 g ammonia is mixed with the same mass of hydrogen chloride. What substances will be present after the reaction has gone to completion, and what will their masses be?

There will remain 5.33 grams of NH3 and there will be produced 14.66 grams NH4Cl

Explanation:

Step 1: Data given

Mass of ammonia = 10.0 grams

Mass of hydrogen chloride = 10.0 grams

Molar mass of ammonia = 17.03 g/mol

Molar mass of hydrogen chloride = 36.46 g/mol

Step 2: the balanced equation

NH3 + HCl → NH4Cl

Step 3: Calculate moles NH3

Moles NH3 = mass NH3 / molar mass NH3

Moles NH3 = 10.0 grams / 17.03 g/mol

Moles NH3 = 0.587 moles

Step 4: Calculate moles HCl

Moles HCl = 10.0 grams / 36.46 g/mol

Moles HCl = 0.274 moles

Step 5: Calculate limiting reactant

For 1 mol NH3 we need 1 mol HCl to produce 1 mol NH4Cl

HCl is the limiting reactant. It will completely be consumed. (0.274 moles)

NH3 is in excess. There will remain 0.587 - 0.274 = 0.313 moles

This is: 0.313 moles * 17.03 = 5.33 grams

Step 6: Calculate moles NH4Cl

For 1 mol NH3 we need 1 mol HCl to produce 1 mol NH4Cl

For 0.274 moles HCl we need 0.274 moles NH4Cl

Step 7: Calculate mass of NH4Cl

Mass NH4Cl = moles NH4Cl * molar mass NH4Cl

Mass NH4Cl = 0.274 moles * 53.49 g/mol

Mass NH4Cl = 14.66 grams

There will remain 5.33 grams of NH3 and there will be produced 14.66 grams NH4Cl